On 2/27/2013 11:40 PM, Tim Williams wrote:
I expect if you reflect the CT secondary stuff (don't forget Lsec) back to
the primary, your answer will appear. Offhand I can't reason out which
sum of L and C makes the resonance, but it's a four pole series-parallel
resonant circuit, analysis should lay it bare.

Are you sure about this? When you say to reflect the secondary back to
the primary that means the primary inductance would be doubled? I
believe the coupling of the two coils means they are one and the same
for the purposes of the circuit analysis, no?

You can't reason which sum of L and C makes the resonance and I can't
either. The calculation is off by about 5% and I can't explain that. I
can explain a null at about 290 kHz. That is the resonance of the
secondary with the secondary capacitance. I can't explain the other
peak at 363 kHz at all. A higher frequency would imply a smaller L
and/or C. How do you combine them to produce that? Consider the two
caps to be in series???

--

Rick

"rickman" wrote in message
...
A higher frequency would imply a smaller L and/or C. How do you combine
them to produce that? Consider the two caps to be in series???

Sure. If you bring the 10p over to the primary, it looks like 10p * (30m
/ 5u), or whatever the ratio was (I don't have it in front of me now), in
parallel with the primary. (I misspoke earlier, you can safely ignore Ls,
because k = 1. There's no flux which is not common to both windings.)

Inductors effectively in parallel also increase the expected resonant
frequency. If you have this,

.. L1
.. +-----UUU--+------+------+
.. | + | | |
.. ( Vsrc ) === C R 3 L2
.. | - | 3
.. | | | |
.. +----------+------+------+
.. _|_ GND

You might expect the resonant frequency is L2 + C, but it's actually (L1
|| L2) = Leq. If L1 is not substantially larger than L2, the resonant
frequency will be pulled higher.

Incidentally, don't forget to include loss components. I didn't see any
explict R on the schematic. I didn't check if you set the LTSpice default
parasitic ESR (cap), or DCR or EPR (coil) on the components. Besides
parasitic losses, your signal is going *somewhere*, and that "where"
consumes power!

The actual transmitter is most certainly not a perfect current source
inductor, nor is the receiver lossless. This simulation has no expression
for radiation in any direction that's not directly between the two
antennas: if all the power transmitted by the current source is reflected
back, even though it's through a 0.1% coupling coefficient, it has to go
somewhere. If it's coming back out the antenna, and it's not being burned
in the "transformer", it's coming back into the transmitter. This is at
odds with reality, where a 100% reflective antenna doesn't magically smoke
a distant transmitter, it simply reflects 99.9% back into space. The
transmitter hardly knows.

In this example, if you set R very large, you'll see ever more voltage on
the output, and ever more current draw from Vsrc. You can mitigate this
by increasing L1 still further, but the point is, if the source and load
(R) aren't matched in some fashion, the power will reflect back to the
transmitter and cause problems (in this case, power reflected back
in-phase causes excessive current draw; in the CCS case, reflected power
in-phase causes minimal voltage generation and little power transmission).

Power is always coming and going somewhere, and if you happen to forget
this fact, it'll reflect back and zap you in the butt sooner or later!

Tim

--
Deep Friar: a very philosophical monk.
Website: http://seventransistorlabs.com

On 2/28/2013 6:40 PM, Tim Williams wrote:
wrote in message
...
A higher frequency would imply a smaller L and/or C. How do you combine
them to produce that? Consider the two caps to be in series???

Sure. If you bring the 10p over to the primary, it looks like 10p * (30m
/ 5u), or whatever the ratio was (I don't have it in front of me now), in
parallel with the primary. (I misspoke earlier, you can safely ignore Ls,
because k = 1. There's no flux which is not common to both windings.)

Reflecting the capacitance through the transformer changes it by the
square of the turns ratio assuming the coupling coefficient is
sufficiently high. I am simulating K at 1.

This is also true for the inductance, but in the opposite manner. So
going from the 25 turn side to the 1 turn side, the effective
capacitance is multiplied by 625 and the effective inductance (or
resistance) is divided by 625. In fact, in LTspice you indicate the
turns ratio by setting the inductance of the two coils by this ratio.

I see now that the reflected secondary capacitance is in parallel with
the primary, rather than in parallel with the primary capacitor. That
explains a lot... I'll have to hit the books to see how to calculate
this new arrangement. I found a very similar circuit in the Radiotron
Designer's Handbook. In section 4.6(iv)E on page 152 they show a
series-parallel combination that only differs in the placement of the
resistance in the parallel circuit. It need to be placed inline with
the inductor... or is placing it parallel correct since this is the
reflected resistance of the secondary? I'll have to cogitate on that a
bit. I'm thinking it would be properly placed inline with the capacitor
in the reflection since it is essentially inline in the secondary.
Either way I expect it will have little impact on the resonant frequency
and I can just toss all the resistances simplifying the math.

I do see one thing immediately. The null in Vcap I see is explained by
the parallel resonance of the secondary cap with the secondary inductor.
If you reflect that cap back to the primary in parallel with the
primary inductor (resonating at the same frequency) it explains the null
in the capacitor C1 voltage I see. C2' (reflected) and L1 make a
parallel resonance with a high impedance dropping the primary cap
current and voltage to a null. This null is calculated accurately.

What I need to do is change the impedance equation from Radiotron to one
indicating the voltage at Vout relative to the input signal. I think I
can do that by treating the circuit as a voltage divider taking the
ratio of the impedance at the input versus the impedance at the primary
coil. No?


Inductors effectively in parallel also increase the expected resonant
frequency. If you have this,

. L1
. +-----UUU--+------+------+
. | + | | |
. ( Vsrc ) === C R 3 L2
. | - | 3
. | | | |
. +----------+------+------+
. _|_ GND

You might expect the resonant frequency is L2 + C, but it's actually (L1
|| L2) = Leq. If L1 is not substantially larger than L2, the resonant
frequency will be pulled higher.

I see, L1 and L2 are in parallel because the impedance of Vsrc is very
low. That is not the circuit I am simulating however. The loop of the
antenna and the loop of the inductor are in series along with the
primary capacitor. I'm not sure what the resistor is intended to
represent, perhaps transformer losses? The resistance of L1 was added
to the simulation model along with the resistance of the secondary coil
which you have not shown... I think. It seems to me you have left out
the tuning capacitor on the primary.


Incidentally, don't forget to include loss components. I didn't see any
explict R on the schematic. I didn't check if you set the LTSpice default
parasitic ESR (cap), or DCR or EPR (coil) on the components. Besides
parasitic losses, your signal is going *somewhere*, and that "where"
consumes power!

The actual transmitter is most certainly not a perfect current source
inductor, nor is the receiver lossless. This simulation has no expression
for radiation in any direction that's not directly between the two
antennas: if all the power transmitted by the current source is reflected
back, even though it's through a 0.1% coupling coefficient, it has to go
somewhere. If it's coming back out the antenna, and it's not being burned
in the "transformer", it's coming back into the transmitter. This is at
odds with reality, where a 100% reflective antenna doesn't magically smoke
a distant transmitter, it simply reflects 99.9% back into space. The
transmitter hardly knows.

Interesting point. My primary goal with this is to simulate the
resonance of the tuning so I can understand how to best tune the
circuit. In many of the simulations I run the Q ends up being high
enough that a very small drift in the parasitic capacitance on the
secondary detunes the antenna and drops the signal level. It sounds
like there are other losses that will bring the Q much lower.

I would also like to have some idea of the signal strength to expect. My
understanding is that the radiation resistance of loop antennas is
pretty low. So not much energy will be radiated out. No?

You make it sound as if in the simulation, even with a small coupling
coefficient all the energy from antenna inductor will still couple back
into the transmitter inductor regardless of the K value. Do I
misunderstand you? It seems to result in the opposite, minimizing this
back coupling. Or are you saying that the simulation needs to simulate
the radiation resistance to show radiated losses?


In this example, if you set R very large, you'll see ever more voltage on
the output, and ever more current draw from Vsrc. You can mitigate this
by increasing L1 still further, but the point is, if the source and load
(R) aren't matched in some fashion, the power will reflect back to the
transmitter and cause problems (in this case, power reflected back
in-phase causes excessive current draw; in the CCS case, reflected power
in-phase causes minimal voltage generation and little power transmission).

Power is always coming and going somewhere, and if you happen to forget
this fact, it'll reflect back and zap you in the butt sooner or later!

Tim

Actually, my goal was to build the receiver and I realized that my
design would require the largest signal I could get from the antenna. I
never realized I would end up having to learn quite so much about
antenna design.

I've been planning to create a PCB with lots of options so I can test a
number of configurations. Nothing about the simulation makes me doubt
the utility of this idea.

One thing that continues to bug me is that nothing I have seen gives me
a hint on how to factor in the distributed capacitance of the antenna
shield. I am using RG6 with 16 pF/Ft and likely will end up with 100
foot of coax total. At some point I'll just have to make some
measurements and see what the real world does.

--

Rick

You'll be much better off simply using the conventional radio approach
than trying to simulate everything, especially when circuit equivalents
are nebulous like this.

After all, if you can't quite tell what it *should* look like, how would
you know if you could implement your model once you've found a
satisfactory result?

What kind of antenna are you looking at, loop? The first thing to know
about a loop is, if it's a very small loop (I'm guessing, at this
frequency, it is), its radiation resistance is very low, meaning, you can
treat it as a nearly pure inductance (Q 10 I think is typical), and its
bandwidth (even with a matched load) will be correspondingly narrow.

The nature of the incoming signal could be modeled as a voltage or current
source; how doesn't really matter, because it isn't really either, it's a
power source that couples in. Again, you don't have voltage without
current and vice versa, it's all about power flow, and the matching that
allows the power to flow.

Since the loop is inductive, your first priority is to resonate it with a
capacitor at the desired frequency. This will require a very precise
value, and even for a single frequency, may require a variable capacitor
to account for manufacturing tolerances. In the AM BCB, a Q of 10 gets
you 50-160kHz bandwidth, so you only get a few channels for any given
tuning position. And if the Q is higher, you get even fewer.

Now that you've got a high Q resonant tank, you can do two things: couple
into the voltage across the capacitor, or the current through the
inductor. You need only a small fraction of either, because the Q is
still going to be large. This can be arranged with a voltage divider
(usually the capacitor is split into a huge hunk and a small variable
part, e.g., 300pF variable + 10nF, output from across the 10nF), a
transformer (a potential transformer across the cap, or a current
transformer in series with the inductor), an inductive pickup (the big
loop carries lots of volts, but you only need a few, so a much smaller
loop can be placed inside the big loop), an impractically large inductor
(like in my example circuit, which models radiation resistance as a
parallel equivalent), etc. Whatever the case, you need to match
transmission line impedance (e.g., 50 ohms) to radiation resistance
(whichever series or parallel equivalent you have).

Once you get the signal into a transmission line, with a reasonable match
(Z ~= Z_line, or alternately, SWR ~= 1), you can do whatever you want with
it. Put it into an amplifier (don't forget to match it, too), etc. Yes,
you're going to have funny behavior at other frequencies, and if you're
concerned about those frequencies, you'll have to choose the coupling
circuit and adjustable (or selectable) components accordingly. But for
the most part, you completely ignore any frequency that you aren't tuning
for, usually enforcing that concept by inserting filters to reject any
stragglers.

Example: suppose you have a loop of 5uH and need to tune it to 500kHz. It
has a reactance of 15.7 ohms. Suppose further it has Q = 20. The ESR
(not counting DCR and skin effect) is X_L / Q, or 0.78 ohms; alternately,
the EPR is X_L * Q, or 314 ohms. The capacitor required is 20.3nF. If we
use a current transformer to match to a 50 ohm line, it needs an impedance
ratio of 1:64, or a turns ratio of 1:8. If we use a voltage transformer,
it's of course 8:1. (A capacitor divider is unsuitable for resonant
impedances less than line impedance, since it can only divide the
impedance down. If the inductance were a lot larger, it could be used.)
To a rough approximation, a smaller inductive loop, of 1/8 diameter of the
larger, I think, would also work.

Tim

--
Deep Friar: a very philosophical monk.
Website: http://seventransistorlabs.com

"rickman" wrote in message
...
On 2/28/2013 6:40 PM, Tim Williams wrote:
wrote in message
...
A higher frequency would imply a smaller L and/or C. How do you
combine
them to produce that? Consider the two caps to be in series???

Sure. If you bring the 10p over to the primary, it looks like 10p *
(30m
/ 5u), or whatever the ratio was (I don't have it in front of me now),
in
parallel with the primary. (I misspoke earlier, you can safely ignore
Ls,
because k = 1. There's no flux which is not common to both windings.)

Reflecting the capacitance through the transformer changes it by the
square of the turns ratio assuming the coupling coefficient is
sufficiently high. I am simulating K at 1.

This is also true for the inductance, but in the opposite manner. So
going from the 25 turn side to the 1 turn side, the effective
capacitance is multiplied by 625 and the effective inductance (or
resistance) is divided by 625. In fact, in LTspice you indicate the
turns ratio by setting the inductance of the two coils by this ratio.

I see now that the reflected secondary capacitance is in parallel with
the primary, rather than in parallel with the primary capacitor. That
explains a lot... I'll have to hit the books to see how to calculate
this new arrangement. I found a very similar circuit in the Radiotron
Designer's Handbook. In section 4.6(iv)E on page 152 they show a
series-parallel combination that only differs in the placement of the
resistance in the parallel circuit. It need to be placed inline with
the inductor... or is placing it parallel correct since this is the
reflected resistance of the secondary? I'll have to cogitate on that a
bit. I'm thinking it would be properly placed inline with the capacitor
in the reflection since it is essentially inline in the secondary.
Either way I expect it will have little impact on the resonant frequency
and I can just toss all the resistances simplifying the math.

I do see one thing immediately. The null in Vcap I see is explained by
the parallel resonance of the secondary cap with the secondary inductor.
If you reflect that cap back to the primary in parallel with the primary
inductor (resonating at the same frequency) it explains the null in the
capacitor C1 voltage I see. C2' (reflected) and L1 make a parallel
resonance with a high impedance dropping the primary cap current and
voltage to a null. This null is calculated accurately.

What I need to do is change the impedance equation from Radiotron to one
indicating the voltage at Vout relative to the input signal. I think I
can do that by treating the circuit as a voltage divider taking the
ratio of the impedance at the input versus the impedance at the primary
coil. No?


Inductors effectively in parallel also increase the expected resonant
frequency. If you have this,

. L1
. +-----UUU--+------+------+
. | + | | |
. ( Vsrc ) === C R 3 L2
. | - | 3
. | | | |
. +----------+------+------+
. _|_ GND

You might expect the resonant frequency is L2 + C, but it's actually
(L1
|| L2) = Leq. If L1 is not substantially larger than L2, the resonant
frequency will be pulled higher.

I see, L1 and L2 are in parallel because the impedance of Vsrc is very
low. That is not the circuit I am simulating however. The loop of the
antenna and the loop of the inductor are in series along with the
primary capacitor. I'm not sure what the resistor is intended to
represent, perhaps transformer losses? The resistance of L1 was added
to the simulation model along with the resistance of the secondary coil
which you have not shown... I think. It seems to me you have left out
the tuning capacitor on the primary.


Incidentally, don't forget to include loss components. I didn't see
any
explict R on the schematic. I didn't check if you set the LTSpice
default
parasitic ESR (cap), or DCR or EPR (coil) on the components. Besides
parasitic losses, your signal is going *somewhere*, and that "where"
consumes power!

The actual transmitter is most certainly not a perfect current source
inductor, nor is the receiver lossless. This simulation has no
expression
for radiation in any direction that's not directly between the two
antennas: if all the power transmitted by the current source is
reflected
back, even though it's through a 0.1% coupling coefficient, it has to
go
somewhere. If it's coming back out the antenna, and it's not being
burned
in the "transformer", it's coming back into the transmitter. This is
at
odds with reality, where a 100% reflective antenna doesn't magically
smoke
a distant transmitter, it simply reflects 99.9% back into space. The
transmitter hardly knows.

Interesting point. My primary goal with this is to simulate the
resonance of the tuning so I can understand how to best tune the
circuit. In many of the simulations I run the Q ends up being high
enough that a very small drift in the parasitic capacitance on the
secondary detunes the antenna and drops the signal level. It sounds
like there are other losses that will bring the Q much lower.

I would also like to have some idea of the signal strength to expect. My
understanding is that the radiation resistance of loop antennas is
pretty low. So not much energy will be radiated out. No?

You make it sound as if in the simulation, even with a small coupling
coefficient all the energy from antenna inductor will still couple back
into the transmitter inductor regardless of the K value. Do I
misunderstand you? It seems to result in the opposite, minimizing this
back coupling. Or are you saying that the simulation needs to simulate
the radiation resistance to show radiated losses?


In this example, if you set R very large, you'll see ever more voltage
on
the output, and ever more current draw from Vsrc. You can mitigate
this
by increasing L1 still further, but the point is, if the source and
load
(R) aren't matched in some fashion, the power will reflect back to the
transmitter and cause problems (in this case, power reflected back
in-phase causes excessive current draw; in the CCS case, reflected
power
in-phase causes minimal voltage generation and little power
transmission).

Power is always coming and going somewhere, and if you happen to forget
this fact, it'll reflect back and zap you in the butt sooner or later!

Tim

Actually, my goal was to build the receiver and I realized that my
design would require the largest signal I could get from the antenna. I
never realized I would end up having to learn quite so much about
antenna design.

I've been planning to create a PCB with lots of options so I can test a
number of configurations. Nothing about the simulation makes me doubt
the utility of this idea.

One thing that continues to bug me is that nothing I have seen gives me
a hint on how to factor in the distributed capacitance of the antenna
shield. I am using RG6 with 16 pF/Ft and likely will end up with 100
foot of coax total. At some point I'll just have to make some
measurements and see what the real world does.

--

Rick

On 3/1/2013 8:53 PM, Tim Williams wrote:
You'll be much better off simply using the conventional radio approach
than trying to simulate everything, especially when circuit equivalents
are nebulous like this.

I don't know what you mean by the "conventional radio approach".


After all, if you can't quite tell what it *should* look like, how would
you know if you could implement your model once you've found a
satisfactory result?

I was simulating a specific circuit for a specific purpose. I got the
answer I was looking for.


What kind of antenna are you looking at, loop? The first thing to know
about a loop is, if it's a very small loop (I'm guessing, at this
frequency, it is), its radiation resistance is very low, meaning, you can
treat it as a nearly pure inductance (Q 10 I think is typical), and its
bandwidth (even with a matched load) will be correspondingly narrow.

Yes, I plan to use a shielded loop. I have found some contradictory
info on the effectiveness of the "shield". One reference seems to have
measurements that show it is primarily E-field coupled in the longer
distance portion of the near-field.

I am aware of the low radiation resistance and have not included that
factor in my simulation. The Q of just the antenna loop is around 100
as calculated from the ratio of reactance to resistance.


The nature of the incoming signal could be modeled as a voltage or current
source; how doesn't really matter, because it isn't really either, it's a
power source that couples in. Again, you don't have voltage without
current and vice versa, it's all about power flow, and the matching that
allows the power to flow.

A friend in a loop antenna Yahoo group suggested the use of the
transformer coupling with a low k to model the signal reception.


Since the loop is inductive, your first priority is to resonate it with a
capacitor at the desired frequency. This will require a very precise
value, and even for a single frequency, may require a variable capacitor
to account for manufacturing tolerances. In the AM BCB, a Q of 10 gets
you 50-160kHz bandwidth, so you only get a few channels for any given
tuning position. And if the Q is higher, you get even fewer.

Yes, that is loop antenna 101 I think. It was when I added a coupling
transformer with 100:1 turns ratio that I was told I needed to consider
the parasitics. I have found it is not useful to go much above 25 or
33:1 on the turns ratio. I am receiving a single frequency, 60 kHz.
There is no need for a wide bandwidth. Ultimately, I prefer a Q of
100 for the higher gain. If it gets too high, the off tuning by
variations (drift) in the parasitic capacitance affects the antenna gain
appreciably.


Now that you've got a high Q resonant tank, you can do two things: couple
into the voltage across the capacitor, or the current through the
inductor. You need only a small fraction of either, because the Q is
still going to be large. This can be arranged with a voltage divider
(usually the capacitor is split into a huge hunk and a small variable
part, e.g., 300pF variable + 10nF, output from across the 10nF), a
transformer (a potential transformer across the cap, or a current
transformer in series with the inductor), an inductive pickup (the big
loop carries lots of volts, but you only need a few, so a much smaller
loop can be placed inside the big loop), an impractically large inductor
(like in my example circuit, which models radiation resistance as a
parallel equivalent), etc. Whatever the case, you need to match
transmission line impedance (e.g., 50 ohms) to radiation resistance
(whichever series or parallel equivalent you have).

Transmission line? What transmission line? The antenna is directly
connected to the receiver which has a very high input impedance. Why do
I need to consider radiation resistance? I have not read that anywhere.


Once you get the signal into a transmission line, with a reasonable match
(Z ~= Z_line, or alternately, SWR ~= 1), you can do whatever you want with
it. Put it into an amplifier (don't forget to match it, too), etc. Yes,
you're going to have funny behavior at other frequencies, and if you're
concerned about those frequencies, you'll have to choose the coupling
circuit and adjustable (or selectable) components accordingly. But for
the most part, you completely ignore any frequency that you aren't tuning
for, usually enforcing that concept by inserting filters to reject any
stragglers.

Example: suppose you have a loop of 5uH and need to tune it to 500kHz. It
has a reactance of 15.7 ohms. Suppose further it has Q = 20. The ESR
(not counting DCR and skin effect) is X_L / Q, or 0.78 ohms; alternately,
the EPR is X_L * Q, or 314 ohms. The capacitor required is 20.3nF. If we
use a current transformer to match to a 50 ohm line, it needs an impedance
ratio of 1:64, or a turns ratio of 1:8. If we use a voltage transformer,
it's of course 8:1. (A capacitor divider is unsuitable for resonant
impedances less than line impedance, since it can only divide the
impedance down. If the inductance were a lot larger, it could be used.)
To a rough approximation, a smaller inductive loop, of 1/8 diameter of the
larger, I think, would also work.

I'm not familiar with the concept of voltage transformer vs. current
transformer. How do you mean that?

How did you get the 1:64 impedance ratio and the 1:8 turns ratio? I
don't follow that. Are you saying the line impedance should match the
ESR? Why exactly would it need to match the ESR?

--

Rick

"rickman" wrote in message
...
Yes, I plan to use a shielded loop. I have found some contradictory
info on the effectiveness of the "shield". One reference seems to have
measurements that show it is primarily E-field coupled in the longer
distance portion of the near-field.

I trust this resource:
http://vk1od.net/antenna/shieldedloop/
He's got gobs of analytical articles.

Yes, that is loop antenna 101 I think. It was when I added a coupling
transformer with 100:1 turns ratio that I was told I needed to consider
the parasitics. I have found it is not useful to go much above 25 or
33:1 on the turns ratio. I am receiving a single frequency, 60 kHz.
There is no need for a wide bandwidth. Ultimately, I prefer a Q of
100 for the higher gain. If it gets too high, the off tuning by
variations (drift) in the parasitic capacitance affects the antenna gain
appreciably.

High Q isn't the goal, high radiation resistance is -- the bigger the
loop, the better it couples with free space, until it's a wave length
around.

You can go ahead and make a teeny coil out of polished silver litz wire,
and push the Q up into the hundreds, but all you'll see is internal
resistance, hardly anything attributable to actual radiation. Since the
losses dominate over radiation, it makes a crappy antenna. But you know
that from looking at it -- it's a tiny lump, of course it's not going to
see the outside world.

It is true, however, that a small coil, with low losses, will have low
noise. AM radios rely on this, which is how they get away with tiny hunks
of ferrite for picking up radio.

Of course, it doesn't hurt that AM stations are 50kW or so, to push over
atmospheric noise.

Transmission line? What transmission line? The antenna is directly
connected to the receiver which has a very high input impedance. Why do
I need to consider radiation resistance? I have not read that
anywhere.

Ok, then you can merge the matching transformer, transmission line and
receiver input transformer into one -- an even larger stepup into whatever
impedance it's looking at (what's "very high", kohms? Mohms?) will get you
that much more SNR.

I'm not familiar with the concept of voltage transformer vs. current
transformer. How do you mean that?

Current transformer measures current (its winding is in series), potential
transformer measures voltage (in parallel).

How did you get the 1:64 impedance ratio and the 1:8 turns ratio? I
don't follow that. Are you saying the line impedance should match the
ESR? Why exactly would it need to match the ESR?

ESR (and Q) measured on the coil corresponds to radiation resistance
(series equivalent) *plus* internal losses (also series equivalent). You
can't separate the two components, so you can only get the best power
match by the good old impedance theorem.

~1:64 is 50 ohm / 0.78 ohm, and N2/N1 = sqrt(Z2/Z1), or 8:1 turns ratio.

Tim

--
Deep Friar: a very philosophical monk.
Website: http://seventransistorlabs.com

On 3/6/2013 8:13 PM, Tim Williams wrote:
wrote in message
...
Yes, I plan to use a shielded loop. I have found some contradictory
info on the effectiveness of the "shield". One reference seems to have
measurements that show it is primarily E-field coupled in the longer
distance portion of the near-field.

I trust this resource:
http://vk1od.net/antenna/shieldedloop/
He's got gobs of analytical articles.

Yes, I've seen this page. Thanks.


Yes, that is loop antenna 101 I think. It was when I added a coupling
transformer with 100:1 turns ratio that I was told I needed to consider
the parasitics. I have found it is not useful to go much above 25 or
33:1 on the turns ratio. I am receiving a single frequency, 60 kHz.
There is no need for a wide bandwidth. Ultimately, I prefer a Q of
100 for the higher gain. If it gets too high, the off tuning by
variations (drift) in the parasitic capacitance affects the antenna gain
appreciably.

High Q isn't the goal, high radiation resistance is -- the bigger the
loop, the better it couples with free space, until it's a wave length
around.

I'm not clear on why you keep referring to radiation resistance for a
receiving antenna. Does this result in a larger received signal? I am
concerned with maximizing the voltage at the input to the receiver.


You can go ahead and make a teeny coil out of polished silver litz wire,
and push the Q up into the hundreds, but all you'll see is internal
resistance, hardly anything attributable to actual radiation. Since the
losses dominate over radiation, it makes a crappy antenna. But you know
that from looking at it -- it's a tiny lump, of course it's not going to
see the outside world.

I have no idea why you are talking about Litz wire and tiny coils. I
never said I was looking to maximize the Q. I said I wanted a Q of over
100. I should have said, slightly over 100. A higher Q clearly does
increase the voltage on the input in my simulations. Is there something
wrong with my simulations?


It is true, however, that a small coil, with low losses, will have low
noise. AM radios rely on this, which is how they get away with tiny hunks
of ferrite for picking up radio.

Of course, it doesn't hurt that AM stations are 50kW or so, to push over
atmospheric noise.

Transmission line? What transmission line? The antenna is directly
connected to the receiver which has a very high input impedance. Why do
I need to consider radiation resistance? I have not read that
anywhere.

Ok, then you can merge the matching transformer, transmission line and
receiver input transformer into one -- an even larger stepup into whatever
impedance it's looking at (what's "very high", kohms? Mohms?) will get you
that much more SNR.

Yes, a higher stepup ratio gets larger signal up to a point. That point
is determined by the parasitic capacitance of the receiver input. That
capacitance is reflected back through the transformer and affects the
antenna tuning. In my simulations it creates a filter with two resonances.


I'm not familiar with the concept of voltage transformer vs. current
transformer. How do you mean that?

Current transformer measures current (its winding is in series), potential
transformer measures voltage (in parallel).

Series and parallel with what? I'm not following this. I have trouble
with series and parallel resonance, but I'm starting to get the concept.
Sometimes it is hard to tell how a circuit is being stimulated.


How did you get the 1:64 impedance ratio and the 1:8 turns ratio? I
don't follow that. Are you saying the line impedance should match the
ESR? Why exactly would it need to match the ESR?

ESR (and Q) measured on the coil corresponds to radiation resistance
(series equivalent) *plus* internal losses (also series equivalent). You
can't separate the two components, so you can only get the best power
match by the good old impedance theorem.

Internal losses of what? How do you determine the internal losses?


~1:64 is 50 ohm / 0.78 ohm, and N2/N1 = sqrt(Z2/Z1), or 8:1 turns ratio.

Ok, so you were matching the hypothetical ESR to the hypothetical line
impedance.

--

Rick

On 6.3.13 9:00 , rickman wrote:

Since the loop is inductive, your first priority is to resonate it with a
capacitor at the desired frequency. This will require a very precise
value, and even for a single frequency, may require a variable capacitor
to account for manufacturing tolerances. In the AM BCB, a Q of 10 gets
you 50-160kHz bandwidth, so you only get a few channels for any given
tuning position. And if the Q is higher, you get even fewer.

Yes, that is loop antenna 101 I think. It was when I added a coupling
transformer with 100:1 turns ratio that I was told I needed to consider
the parasitics. I have found it is not useful to go much above 25 or
33:1 on the turns ratio. I am receiving a single frequency, 60 kHz.
There is no need for a wide bandwidth. Ultimately, I prefer a Q of
100 for the higher gain. If it gets too high, the off tuning by
variations (drift) in the parasitic capacitance affects the antenna gain
appreciably.

Please note that high Q will destroy the modulation sidebands on
the signal you're listening to.

In aviation, there are non-directional beacons which are transmitting
in a frequency around 300 kHz (1 km wavelength). The antennas cannot
obviously be of efficient length (250 m / 800 ft), so they are short
(20 m / 70 ft) force-tuned to the transmitting frequency. This creates
so high Q that the identification modulation sidebands for the customary
1050 Hz audio do not fit in, and the ID is modulated using 400 Hz audio.

--

Tauno Voipio, avionics engineer (also OH2UG)

On 3/7/2013 2:15 AM, Tauno Voipio wrote:
On 6.3.13 9:00 , rickman wrote:

Since the loop is inductive, your first priority is to resonate it
with a
capacitor at the desired frequency. This will require a very precise
value, and even for a single frequency, may require a variable capacitor
to account for manufacturing tolerances. In the AM BCB, a Q of 10 gets
you 50-160kHz bandwidth, so you only get a few channels for any given
tuning position. And if the Q is higher, you get even fewer.

Yes, that is loop antenna 101 I think. It was when I added a coupling
transformer with 100:1 turns ratio that I was told I needed to consider
the parasitics. I have found it is not useful to go much above 25 or
33:1 on the turns ratio. I am receiving a single frequency, 60 kHz.
There is no need for a wide bandwidth. Ultimately, I prefer a Q of
100 for the higher gain. If it gets too high, the off tuning by
variations (drift) in the parasitic capacitance affects the antenna gain
appreciably.

Please note that high Q will destroy the modulation sidebands on
the signal you're listening to.

I appreciate the advice from everyone, but much of it is not in the
proper context and way off target. "High" Q is how high? Where are the
modulation sidebands? My point is that I have already considered this.
The modulation sidebands of this signal are on the order of low 10's
of Hz. This signal is modulated at a 1 bit per second rate. I will be
demodulating a 30 Hz sample rate. So a bandwidth of 100 Hz is plenty
which corresponds to a Q of around 500.

I said I was looking for a Q over 100, maybe I should have said a Q of a
bit over 100. By the time it gets to 300 it is to peaky to hold a tune
setting. That is the problem I am concerned with.


In aviation, there are non-directional beacons which are transmitting
in a frequency around 300 kHz (1 km wavelength). The antennas cannot
obviously be of efficient length (250 m / 800 ft), so they are short
(20 m / 70 ft) force-tuned to the transmitting frequency. This creates
so high Q that the identification modulation sidebands for the customary
1050 Hz audio do not fit in, and the ID is modulated using 400 Hz audio.

Ok, but that is nothing like my application, receiving WWVB.

--

Rick